Spivak's proof of Inverse Function Theorem I am having trouble with Spivak's proof of the Inverse Function Theorem in his Calculus on Manifolds:
2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and det $f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$
Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$...
How is the theorem true for $f$ if it is true for $\lambda^{-1}\circ f$?
 A: Suppose the theorem holds for all functions $f$ such that $Df(a) = I$. What Spivak has shown is that for $\lambda = Df(a)$, $D(\lambda^{-1} \circ f(a)) = I$. By the assumption above, the theorem must be true for $\lambda^{-1}\circ f$. That means there exists $g$ which is the continuously differentiable inverse of $\lambda^{-1} \circ f$. This implies that $(g \circ \lambda^{-1}) \circ f = I$. Now $g \circ \lambda^{-1}$ is continuously differentiable, and so the theorem is true for $f$.
A: $\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$ is a bijection and $\lambda$ and $\lambda^{-1}$ are both continuously differentiable.
Note that $\lambda'(z) = \lambda$ for all $z \in \mathbb{R}$.
Let $g = \lambda^{-1}\circ f$.
Suppose the theorem is true for $g$.
Then there is an open set $V'$ containing $a$ and an open set $W'$ containing $g(a)$ such that $g:V'\to W'$ has a continuous inverse $g^{-1}:W'\to V'$ which is differentiable and for all $y\in W'$ satisfies $$(g^{-1})'(y) = [g'(g^{-1}(y))]^{-1}$$
Then $\lambda(W')$ is open and $f = \lambda\circ g:V'\to \lambda(W')$ has a continuous inverse $g^{-1}\circ \lambda^{-1}:\lambda(W')\to V'$.
By the chain rule, for all $z \in \lambda(W')$, $(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$
A: Let $f: \mathbb{R}^n\to\mathbb{R}^n$ be a continuously differentiable function in an open set containing $a$, and det $f'(a)\neq 0$.
Let $\lambda := Df(a)$.
Then $\lambda$ is a non-singular linear transformation because $\det f'(a) \neq 0$.
Let $g := \lambda^{-1} \circ f$ and $f = (f^1, \cdots, f^n)$ and $g = (g^1, \cdots, g^n)$.
Then we can write $g^i(x)$ as $g^i(x) = \mu_{i 1} f^1(x) + \cdots + \mu_{i n} f^n(x)$ for some real numbers $\mu_{i 1}, \cdots, \mu_{i n}$.
Then each $g^i$ is a continuously differentiable function in an open set containing $a$ because each $f^i$ is a continuously differentiable function in an open set containing $a$.
So, $g:\mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$.
$D(\lambda^{-1} \circ f)(a) = D\lambda^{-1}(f(a)) \circ Df(a) = \lambda^{-1} \circ Df(a) = I$.
So, $g'(a) = I_n$.
So, $\det g'(a)=1\neq 0$.
Then by the Spivak's proof in the text, there is an open set $V$ containing $a$ and an open set $W$ containing $g(a)$ such that $g:V\to W$ has a continuous inverse $g^{-1} : W \to V$ which is differentiable and for all $y\in W$ satisfies $$(g^{-1})^{'}(y)=[g^{'}(f^{-1}(y))]^{-1}.$$
$f = \lambda \circ g$.
The image of $\lambda : W \to \mathbb{R}^n$ is $\lambda (W)$.
$\lambda : W \to \lambda (W)$ is injective because $\lambda : \mathbb{R}^n \to \mathbb{R}^n$ is non-singular.
So, $\lambda : W \to \lambda (W)$ is bijective.
So $f=\lambda\circ g : V \to \lambda (W)$ is bijective.
$\lambda W$ is an open set. A proof of this fact is the following:
Let $\lambda(w) \in \lambda (W)$.
By the problem 1-10 on p.5, there exists a positive real number $M$ such that $|\lambda^{-1} (h)| \leq M |h|$ for any $h \in \mathbb{R}^n$.
Because $W$ is open, there exists $\epsilon > 0$ such that for any $w' \in \mathbb{R}^n$, if $|w' - w| < \epsilon$,  then $w' \in W$.
Let $u'$ be an arbitrary element of $\mathbb{R}^n$ such that $|u' - \lambda w| < \frac{\epsilon}{M}$.
Then, $|\lambda^{-1} (u') - w| = |\lambda^{-1} (u') - \lambda^{-1} (\lambda w)| = |\lambda^{-1} (u' - \lambda w)| \leq M |u' - \lambda w| < M \frac{\epsilon}{M} = \epsilon$.
So, $\lambda^{-1} (u') \in W$.
So, $u' = \lambda(\lambda^{-1}(u')) \in \lambda (W)$.
So, $\lambda (W)$ is open.
Now $f : V \to \lambda (W)$ is a continuously differentiable bijective function from an open set $V$ to an open set $\lambda (W)$.
$f^{-1} = g^{-1} \circ \lambda^{-1}$.
$g^{-1}$ is differentiable on $W$ and $\lambda^{-1}$ is differentiable on $\lambda (W)$.
So, $f^{-1}$ is differentiable on $\lambda (W)$ by Chain Rule.
Let $\lambda^{'}(x)=A$.
Then, \begin{align*}
(f^{-1})^{'}(y)
&=
(g^{-1}\circ\lambda^{-1})^{'}(y)
=
(g^{-1})^{'}(\lambda^{-1}(y))\cdot(\lambda^{-1})^{'}(y)\\
&=[g^{'}(g^{-1}(\lambda^{-1}(y)))]^{-1}\cdot A^{-1}
=[g^{'}(f^{-1}(y))]^{-1}\cdot A^{-1}\\
&=[A\cdot g^{'}(f^{-1}(y))]^{-1}
=[(\lambda\circ g)^{'}(f^{-1}(y))]^{-1}\\
&=[f^{'}(f^{-1}(y))]^{-1}.
\end{align*}
